How not to copy the file twice when getting the hash?

Question

go

The file is uploaded to the server and stored in multipart.File, then you must first get its hash, and only then create the file on disk. The problem is that the io.Copy function is called twice. First, in the hash getting function, and then when copying from the form to a file on disk. How can this be avoided?

i.e. the first time it is called here

hash, err := GetFileHashMD5(fileFromForm)

// это уже в другом пакете
func GetFileHashMD5(file io.Reader) (hashed string, err error) {
  hash := md5.New()
  if _, err := io.Copy(hash, file); err != nil {
    return "", err
  }
  hashed = hex.EncodeToString(hash.Sum(nil))
  return
}

and then again here, after receiving the hash and creating an empty file on the disk,
_, err = io.Copy(createdFile, fileFromForm)
I tried to immediately write the file from the form to the []byte variable, and then pass this byte array to the hashing function, and write it to the file on the disk. But then the file on the disk becomes damaged (does not open), and besides, as I understand it, if you write the file to a variable, it will clog the RAM.

Answer 1

If the file is small and can be read into memory then the algorithm might be:

// вычитываете содержимое при помощи io.Reader'a, например
fh, handler, err := r.FormFile("myFile")
fileBytes, err := io.ReadAll(fh)

// считаете md5
md5sum := md5.Sum(fileBytes)

// сохраняете файл
f, err := os.Create("/file_path/file_name")
defer f.Close()
f.Write(fileBytes)